Pitomath
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- asked 6 months agoVotes
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Giải Toán 11 Ôn tập cuối năm
a) limx→−26−3x2x2+1;
b) limx→2x−3x−2x2−4;
c) limx→2+x2−3x+1x−2;
d) limx→1−x+x2+..+xn−n1−xvới n∈ℕ*;
e) limx→+∞2x−1x+3;
f) limx→−∞x+4x2−12−3x;
g) limx→−∞−2x3+x2−3x+1.
Lời giải:
a)limx→−26−3x2x2+1=6−3.−22.−22+1=123=4b)limx→2x−3x−2x2−4=limx→2x−3x−2x+3x−2x2−4x+3x−2=limx→2x2−3x+2x2−4x+3x−2=limx→2x−2x−1x−2x+2x+3x−2=limx→2x−1x+2x+3x−2=2−12+22+3.2−2=116c)limx→2+x2−3x+1x−2limx→2+x2−3x+1=4−6+1=−1x−2>0limx→2+x−2=0
Do đó: limx→2+x2−3x+1x−2=−∞
d) limx→1−x+x2+..+xn−n1−x
Ta có:
x+x2+…+xn−n1−x=x1−xn1−x−n1−x=x1−xn−n1−x⇒limx→1−x+x2+…+xn−n1−x=limx→1−x1−xn−n1−x
Mà limx→1−x1−xn−n=1(1−1)−n=−n<0
Và limx→1−(1−x)=01−x>0 khi x<1nên limx→1−x1−xn−n1−x=−∞.
e)limx→+∞2x−1x+3=limx→+∞x2−1xx1+3x=limx→+∞2−1x1+3x=2.
f)limx→−∞x+4x2−12−3x=limx→−∞x+|x|4−1x22−3x=limx→−∞x−x4−1x22−3x=limx→−∞x1−4−1x2x2x−3=limx→−∞1−4−1x22x−3=1−4−3=13
g) limx→−∞−2x3+x2−3x+1
=limx→−∞x3−2+1x−3x2+1x3=+∞
Do limx→−∞x3=−∞và limx→−∞−2+1x−3x2+1x3=−2<0.
